Vector spaces, Subspaces
We have vector spaces
\mathbb{R}^n \text{ (scalars are in $\mathbb{R}$) }
\mathbb{C}^n \text{ (scalars are in $\mathbb{C}$) }
And subspaces, subsets of \mathbb{R}^n closed under addition and scalar multiplication.
x_1,x_2 \in S \implies c_1x_1 + c_2x_2 \in S \quad\forall c_1,c_2 \in \mathbb{R}
Fun fact, Ax = b can be solved for x if and only if b \in \text{Col}(A), that meaning b is a linear combination of the columns of A.
Section 3.2
Let A of size m \times n be an arbitrary matrix.
We have
\text{Null}(A) = \{x \in \mathbb{R}^n \mid Ax = 0\}
If \text{Null}(A) is a subspace of \mathbb{R}^n, then linearity applies.
Fact. Ax = 0 if and only if x \perp \text{RowSp}(A).
We say \text{Null}(A) = \text{Row}(A)^\perp. How do we find \text{Null}(A)? Row operations on A don’t affect the null space.
A strategy to find the null space on A is to convert A into reduced row echelon form R. The row space remains the same so the null space remains the same. The column spaces however are entirely different.
Converting to RREF is simple. So I won’t write it down.
Section 3.4
Recall that for vectors v_1,\cdots,v_n in R^n.
- \text{Span}(\{v_1,\ldots,v_n\}) is the set of all linear combinations of v_1 to v_n.
- v_1,\ldots,v_n are linearly dependent if \exists c_1,\ldots,c_n not all zero such that c_1v_1,\ldots,c_nv_n = 0.
- If the vectors are not linearly dependent, then they are linearly independent.
Suppose v_1,\ldots,v_n are linearly independent in \mathbb{R}^m and
w \in \mathbb{R}^m, w \not \in \text{Span}(\{v_1,\ldots,v_n\})
Then \{v_1,\ldots,v_n,w\} is linearly independent.
Suppose v_1,\ldots,v_n are linearly dependent in \mathbb{R}^m. Then for some j
\text{Span}\{v_1,\ldots,v_{j-1},v_{j+1},\ldots,v_n\} = \text{Span}\{v_1,\ldots,v_n\}
Suppose v_1,\ldots,v_n are linearly dependent in \mathbb{R}^m for n>m, then the vectors are linearly dependent.
Set
\begin{align*}
A = \begin{pmatrix}
1 & & 1\\
v_1 & \cdots & v_n\\
1 & & 1
\end{pmatrix}
\end{align*}
Note n>m \implies \text{Null}(A) \not = \varnothing, by row reduction. Thus there exist some Ac = 0, c \not = 0. So the vectors are linearly dependent
Let V be a subspace of \mathbb{R}^m. A basis of V is a linearly independent set of vectors v_1,\ldots,v_r \in V such that their span is equal to the span of V.
- Any linearly independent subset of V can be enlarged to a basis.
- Any subset of V which spans V can be shrunk to a basis.
Suppose \{w_i,\ldots,w_s\} are linearly independent in V. If their span is V, done. If not, take w_{s+1} in V, not in the span. Then w_i,\ldots,w_{s+1} is linearly independent. By Fact 3.4.3, this terminates at some r \leq m.
Given a set S of vectors which span V, let T be a subset of S of minimal size which spans V. S is linearly independent.
Suppose V is a subspace of \mathbb{R}^m. Let
- u_1,\ldots,u_p span V
- w_1,\ldots,u_r are linearly independent in V.
Using only vectors belonging to u, we can enlarge w to a basis.
If w spans V, we are done. If not, there exists some u not in \text{Span}(w). Add u to w, and repeat until w spans V.
Suppose V is a subspace of \mathbb{R}^m, such that w_1,\ldots,w_r is a basis of V. Suppose we also have u_1,\ldots,u_p \in V, p > r. Then u_1,\ldots,u_p are linearly dependent.
We want c's in \mathbb{R}, not all 0, such that u_1c_1 + \cdots + u_pc_p = 0. Because w is a basis, we have
u_1 = w_1A_{11} + \cdots + w_r A_{r1}
u_2 = w_1A_{12} + \cdots + w_r A_{r2}
\vdots
u_p = w_1A_{1p} + \cdots + w_r A_{rp}
Multipliying each u_i by c_i, we get
u_1c_1 + \cdots + u_pc_p = w_1(Ac)_1 + \cdots + w_p(Ac)_p
Because p > r, we have a non empty null space, thus we prove that the c we want exists.
Suppose V is a subspace of \mathbb{R}^m. All bases of V are the same size, which is called the dimension of V.
Suppose we have two bases w and u of size r and p respecitvly. If r > p, then by fact 3.4.6, w would not be a basis. If p < r, then u would not be a basis. Thus r = p.
Section 3.5
Let A be an (m\times n) matrix. Row operations on A do not change the row space. Thus
\text{RowSp(EA) = RowSp(A)}
If E is an invertable matrix. Similarly for column space,
\text{ColSp(AE) = ColSp(A)}.
Each row operation on A merely takes a linear combination of rows and and replaces some row with it. Thus the span of the rows remains the same.
Let E be an (m\times m) invertable matrix, V a subspace of \mathbb{R}^m. Then EV = \{Ev \mid v \in V\} is also a subspace, and E maps bases of V to bases of EV.
Because c_1w_1, + \cdots + c_rw_r = 0, We have Ew = E0 = 0. Thus Ew is linearly independent and a basis.
Row operations do not change the dimension of the column space, nor the span of the column space. The same applies to column operations.
This immeditally follows from 3.5.2. After applying row operations we are left with EA.
\text{ColSp}(Ea) = E\text{ColSp}(A).
The same applies to column operations.
Suppose v_1,\ldots,v_p in \mathbb{R}^\ell have span V.
This immeditally follows from 3.5.2. After applying row operations we are left with EA.
\text{ColSp}(Ea) = E\text{ColSp}(A).
The same applies to column operations.
The row space and column space of any matrix A have the same dimension
Let A_{m\times n} be a linear map A \colon \mathbb{R^n} \to \mathbb{R}^m
Our null space of A has dimension n - r, while our row space has dimension r. Note that
\text{Image}(A) = A(\mathbb{R}^n) = \text{ColSp}(A)
Note that
\text{Rank}(A) = \text{Rank}(A^T)
The dimension of our \text{Null}(A) = n-r, where the basis is our special solutions.
The Nullspace matrix of the checkers matrix is
- A negated rowspace with the standard basis tacked on.